Problem Link :- leetcode-2187
Problem statement
You are given an array time where time[i] denotes the time taken by the ith bus to complete one trip.
Each bus can make multiple trips successively; that is, the next trip can start immediately after completing the current trip. Also, each bus operates independently; that is, the trips of one bus do not influence the trips of any other bus.
You are also given an integer totalTrips, which denotes the number of trips all buses should make in total. Return the minimum time required for all buses to complete at least totalTrips trips.
Each bus can make multiple trips successively; that is, the next trip can start immediately after completing the current trip. Also, each bus operates independently; that is, the trips of one bus do not influence the trips of any other bus.
You are also given an integer totalTrips, which denotes the number of trips all buses should make in total. Return the minimum time required for all buses to complete at least totalTrips trips.
Significance of Problem statement
This problem presents a scenario involving buses and their trip durations, aiming to find the minimum time required for all buses to collectively complete a certain number of trips. Each bus operates independently, and the next trip can start immediately after finishing the current one. The task is to optimize the total time taken by the buses to meet or exceed a specified total number of trips.
In simpler terms, imagine you have a bunch of buses, each taking a certain amount of time for a trip. They can keep making trips one after another, and you're given a goal: a total number of trips all buses should accomplish. The challenge is to figure out the quickest way for these buses to achieve that total number of trips collectively.
The provided Python code suggests a binary search approach, which is a clever way to efficiently narrow down the possibilities and find the optimal solution. It works by setting a time range (from 0 to a very large number) and iteratively narrowing it down until the minimum time that satisfies the total trips requirement is found. The binary search efficiently explores different time scenarios, significantly reducing the time complexity of the solution.
The significance of this problem lies in its real-world applicability. In transportation or logistics, understanding how to optimize the scheduling of multiple independent entities, such as buses, is crucial. This problem challenges programmers to think strategically, leveraging binary search to efficiently explore potential solutions. It's an opportunity for beginners to dive into algorithmic thinking, exploring a scenario where efficiency in scheduling and resource utilization is key.
By solving this problem, beginners not only gain exposure to advanced programming techniques like binary search but also develop a practical understanding of how such algorithms can be applied to solve real-world problems, making their journey in the programming world more engaging and insightful.
In simpler terms, imagine you have a bunch of buses, each taking a certain amount of time for a trip. They can keep making trips one after another, and you're given a goal: a total number of trips all buses should accomplish. The challenge is to figure out the quickest way for these buses to achieve that total number of trips collectively.
The provided Python code suggests a binary search approach, which is a clever way to efficiently narrow down the possibilities and find the optimal solution. It works by setting a time range (from 0 to a very large number) and iteratively narrowing it down until the minimum time that satisfies the total trips requirement is found. The binary search efficiently explores different time scenarios, significantly reducing the time complexity of the solution.
The significance of this problem lies in its real-world applicability. In transportation or logistics, understanding how to optimize the scheduling of multiple independent entities, such as buses, is crucial. This problem challenges programmers to think strategically, leveraging binary search to efficiently explore potential solutions. It's an opportunity for beginners to dive into algorithmic thinking, exploring a scenario where efficiency in scheduling and resource utilization is key.
By solving this problem, beginners not only gain exposure to advanced programming techniques like binary search but also develop a practical understanding of how such algorithms can be applied to solve real-world problems, making their journey in the programming world more engaging and insightful.
What will you learn
This problem teaches us how to efficiently manage buses to achieve a specific number of trips. Imagine you have buses, each taking different times for a trip. They can keep going without waiting, and you're given a total number of trips they should complete together. The challenge is to find the quickest way for all buses to reach or exceed that total number of trips.
The clever part of the solution lies in using a technique called binary search, like searching for a word in a dictionary. The code sets a range of time values (from 0 to a very large number) and narrows it down systematically until it finds the minimum time needed to meet the total trips goal. Binary search makes the process faster and more efficient.
The significance of this problem goes beyond buses and trips; it reflects a real-world scenario in logistics and scheduling. In transportation, like managing buses, optimizing schedules is essential. This problem introduces beginners to algorithmic thinking, helping them understand how to solve complex problems efficiently.
By grappling with this problem, beginners learn how to approach real-world situations programmatically. They gain insights into strategic thinking and how to use advanced techniques like binary search to cut down the time it takes to find solutions. This experience is like a practical puzzle that not only hones programming skills but also highlights the power of algorithms in solving problems with efficiency.
In a nutshell, this problem is a gateway for beginners to dive into the world of efficient scheduling and resource management. It's a step towards mastering algorithms, providing valuable lessons that extend beyond the realm of buses and trips into the broader landscape of problem-solving in the programming universe.
The clever part of the solution lies in using a technique called binary search, like searching for a word in a dictionary. The code sets a range of time values (from 0 to a very large number) and narrows it down systematically until it finds the minimum time needed to meet the total trips goal. Binary search makes the process faster and more efficient.
The significance of this problem goes beyond buses and trips; it reflects a real-world scenario in logistics and scheduling. In transportation, like managing buses, optimizing schedules is essential. This problem introduces beginners to algorithmic thinking, helping them understand how to solve complex problems efficiently.
By grappling with this problem, beginners learn how to approach real-world situations programmatically. They gain insights into strategic thinking and how to use advanced techniques like binary search to cut down the time it takes to find solutions. This experience is like a practical puzzle that not only hones programming skills but also highlights the power of algorithms in solving problems with efficiency.
In a nutshell, this problem is a gateway for beginners to dive into the world of efficient scheduling and resource management. It's a step towards mastering algorithms, providing valuable lessons that extend beyond the realm of buses and trips into the broader landscape of problem-solving in the programming universe.
Easiest Explanation
Imagine we have a bunch of buses and we want to figure out how long it will take them to complete a certain number of trips.
We know how long it takes each bus to complete one trip, and we know that each bus can keep going on trip after trip without stopping.
But we don't know how many trips each bus will need to complete in order to reach our target number of total trips.
So what we can do is try different numbers of trips for each bus, and see how long it takes them to complete those trips. We can start with a maximum number(worst case which is 10^14) of trips, and keep decreasing it until we reach our target number of total trips.
But checking every possible number of trips for each bus would take a really long time! So instead, we can use something called binary search.
Binary search is like a guessing game. We start by guessing a number (let's say, 10), and then we check if that number is too high or too low.
If it's too high, we know that any number higher than 10 is also too high, so we can guess a lower number (let's say, 5).
If it's too low, we know that any number lower than 10 is also too low, so we can guess a higher number (let's say, 15).
We keep guessing and adjusting our range until we find the exact number we're looking for.
In this case, we can use binary search to guess the number of trips each bus should make in order to reach our target number of total trips. We can start by guessing a highest number (10^14), and keep decreasing it until we reach our target number.
Each time we guess a number, we can calculate how long it would take all the buses to complete that number of trips. If it's too high, we know we need to guess a lower number. If it's too low, we know we need to guess a higher number.
We keep guessing and adjusting our range until we find the minimum amount of time it will take for all the buses to complete at least the target number of trips.
I hope that explanation helps! If you find any difficulty in solving, feel free to comment with your doubts.
We know how long it takes each bus to complete one trip, and we know that each bus can keep going on trip after trip without stopping.
But we don't know how many trips each bus will need to complete in order to reach our target number of total trips.
So what we can do is try different numbers of trips for each bus, and see how long it takes them to complete those trips. We can start with a maximum number(worst case which is 10^14) of trips, and keep decreasing it until we reach our target number of total trips.
But checking every possible number of trips for each bus would take a really long time! So instead, we can use something called binary search.
Binary search is like a guessing game. We start by guessing a number (let's say, 10), and then we check if that number is too high or too low.
If it's too high, we know that any number higher than 10 is also too high, so we can guess a lower number (let's say, 5).
If it's too low, we know that any number lower than 10 is also too low, so we can guess a higher number (let's say, 15).
We keep guessing and adjusting our range until we find the exact number we're looking for.
In this case, we can use binary search to guess the number of trips each bus should make in order to reach our target number of total trips. We can start by guessing a highest number (10^14), and keep decreasing it until we reach our target number.
Each time we guess a number, we can calculate how long it would take all the buses to complete that number of trips. If it's too high, we know we need to guess a lower number. If it's too low, we know we need to guess a higher number.
We keep guessing and adjusting our range until we find the minimum amount of time it will take for all the buses to complete at least the target number of trips.
I hope that explanation helps! If you find any difficulty in solving, feel free to comment with your doubts.
class Solution:
def minimumTime(self, time: List[int], totalTrips: int) -> int:
start = 0
end = 10**14 #(10^14) worst case
ans = end #worst case
while start < end:
mid = (start + end) // 2
cnt = 0
for i in time:
cnt += mid // i
if cnt >= totalTrips:
ans = min(ans, mid)
end = mid
else:
start = mid + 1
return ans
class Solution { public: long long minimumTime(vector<int>& time, int totalTrips) { long long start=0; long long end = 1e14;//(10^14) worst case long long ans = end;//worst case while(start<end){ long long mid = (start+end)/2; long long cnt = 0; for(auto i : time){ cnt +=mid/i; } if(cnt >= totalTrips){ ans = min(ans,mid); end = mid; } else start=mid+1; } return ans; } };
Here's how the function works:
The code uses binary search to find the answer quickly.Here's what the code is doing:
1. First, it initializes some variables. start is set to 0, which is the lowest possible number of trips for any bus, and end is set to a really big number (10^14), which is the highest possible number of trips for any bus. ans is also set to end, which is just a starting value.
2. The code then enters a while loop that will keep running until start and end are the same value. This is the binary search part!
3. Inside the loop, the code calculates the mid value by taking the average of start and end. This is the number of trips that each bus will make in this iteration.
4. The code then counts the total number of trips that all the buses will make if they each make mid trips. It does this by looping through each bus in the time array and dividing mid by the time it takes for that bus to complete one trip. This gives us the number of trips that bus can make in mid time.
5. If the total number of trips is greater than or equal to the target number of trips (totalTrips), the code updates ans to be the minimum of ans and mid. This means that if we can make at least totalTrips trips in mid time, we want to remember mid as a possible answer.
6. If the total number of trips is less than totalTrips, the code updates start to be mid + 1. This means that we need to try a higher number of trips for each bus.
7. The while loop repeats, with start and end adjusted according to the result of the previous iteration.
8. Once start and end are the same value, the loop stops, and the code returns the final value of ans, which is the minimum time required for all buses to complete at least totalTrips trips.
That's it! The code uses binary search to quickly find the answer, and it's a very efficient way to solve this kind of problem.
1. First, it initializes some variables. start is set to 0, which is the lowest possible number of trips for any bus, and end is set to a really big number (10^14), which is the highest possible number of trips for any bus. ans is also set to end, which is just a starting value.
2. The code then enters a while loop that will keep running until start and end are the same value. This is the binary search part!
3. Inside the loop, the code calculates the mid value by taking the average of start and end. This is the number of trips that each bus will make in this iteration.
4. The code then counts the total number of trips that all the buses will make if they each make mid trips. It does this by looping through each bus in the time array and dividing mid by the time it takes for that bus to complete one trip. This gives us the number of trips that bus can make in mid time.
5. If the total number of trips is greater than or equal to the target number of trips (totalTrips), the code updates ans to be the minimum of ans and mid. This means that if we can make at least totalTrips trips in mid time, we want to remember mid as a possible answer.
6. If the total number of trips is less than totalTrips, the code updates start to be mid + 1. This means that we need to try a higher number of trips for each bus.
7. The while loop repeats, with start and end adjusted according to the result of the previous iteration.
8. Once start and end are the same value, the loop stops, and the code returns the final value of ans, which is the minimum time required for all buses to complete at least totalTrips trips.
That's it! The code uses binary search to quickly find the answer, and it's a very efficient way to solve this kind of problem.
Conclusion
In conclusion, this problem offers beginners a hands-on journey into the practical side of programming. It's like orchestrating a fleet of buses efficiently to meet a specific goal of total trips. The ingenious use of binary search in the solution introduces learners to a powerful tool for quickly finding optimal solutions. The significance here is not just about buses; it's a sneak peek into the real-world challenges of scheduling and resource management.
By tackling this problem, beginners not only refine their programming skills but also grasp the importance of strategic thinking in solving problems. It's a step towards becoming adept problem-solvers, equipped with tools like binary search that make complex tasks simpler and more manageable. This adventure with buses and trips is a valuable lesson in the art of efficient problem-solving in the programming realm.
By tackling this problem, beginners not only refine their programming skills but also grasp the importance of strategic thinking in solving problems. It's a step towards becoming adept problem-solvers, equipped with tools like binary search that make complex tasks simpler and more manageable. This adventure with buses and trips is a valuable lesson in the art of efficient problem-solving in the programming realm.
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